Let $g$ be a vector-valued function defined by $g(t)=\left(-10\log(3t),-\dfrac{5}{6t}\right)$. Find $g$ 's second derivative $g''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(\dfrac{10}{t^2\ln(10)},-\dfrac{5}{3t^3}\right)$ (Choice B) B $\left(-\dfrac{10}{t^2\ln(3)},-\dfrac{5}{6t^3}\right)$ (Choice C) C $\left(-\dfrac{10}{9t^2\ln(10)},-\dfrac{5}{3t^3}\right)$ (Choice D) D $\left(-\dfrac{10}{t\ln(10)},\dfrac{5}{6t^2}\right)$
Answer: We are asked to find the second derivative of $g$. This means we need to differentiate $g$ twice. In other words, we differentiate $g$ once to find $g'$, and then differentiate $g'$ (which is a vector-valued function as well) to find $g''$. Recall that $g(t)=\left(-10\log(3t),-\dfrac{5}{6t}\right)$. Therefore, $g'(t)=\left(-\dfrac{10}{t\ln(10)},\dfrac{5}{6t^2}\right)$. Now let's differentiate $g'(t)=\left(-\dfrac{10}{t\ln(10)},\dfrac{5}{6t^2}\right)$ to find $g''$. $g''(t)=\left(\dfrac{10}{t^2\ln(10)},-\dfrac{5}{3t^3}\right)$ In conclusion, $g''(t)=\left(\dfrac{10}{t^2\ln(10)},-\dfrac{5}{3t^3}\right)$.